CONVEX SETS IN HILBERT SPACE
F. S. DE BLASI AND N. V. ZHIVKOV Received 22 November 2003
For a nonempty separable convex subsetXof a Hilbert spaceH(Ω), it is typical (in the sense of Baire category) that a bounded closed convex setC⊂H(Ω) defines anm-valued metric antiprojection (farthest point mapping) at the points of a dense subset ofX, when- evermis a positive integer such thatm≤dimX+ 1.
1. Introduction
Baire category techniques are known to be a powerful tool in the investigation of the convex sets. Their use, which goes back to the fundamental contribution of Klee [17], has permitted to discover several interesting unexpected properties of convex sets (see Gruber [14], Schneider [23], Zamfirescu [25]). A survey of this area of research and additional bibliography can be found in [15,27].
In the present paper, we consider some geometric properties of typical (in the sense of the Baire categories) nonempty bounded closed convex sets contained in a separable real Hilbert space. It will be shown in the typical case, for a closed convex and bounded setCand an integerm, that there is a dense subsetDof the Hilbert spaceHsuch that the farthest point mapping generated byCis precisely m-valued at the points ofD. A result of this type was recently obtained in [5], for typical nonempty compact convex sets.
However, the approach of [5] cannot be adopted here for, in absence of compactness, the antiprojection mapping could have empty images. To overcome this difficulty we will use some ideas from [28], developed in the framework of the metric projections.
Throughout, H(Ω) is a Hilbert space over the field of real numbers Rwhose ele- ments are mappingsx:Ω→Rwith countably many nonzero values and convergent sums
ω∈Ωxω2. We often prefer to denoteH(Ω) byH. It is assumed always in the paper that dimH(Ω)≥2. As usual, the inner product and the norm are denoted by·,·and| · |.
For a nonempty bounded setM⊂H, the function f(x,M)=sup{|x−z|:z∈M}is the farthest distance function, and the set-valued mapping
Q(x,M)=
y∈M:|x−y| = f(x,M) (1.1) is calledmetric antiprojectionorfarthest point mapping.
Copyright©2005 Hindawi Publishing Corporation Abstract and Applied Analysis 2005:4 (2005) 423–436 DOI:10.1155/AAA.2005.423
LetᏯbe the complete metric space of all nonempty bounded closed convex subsets ofH, endowed with the Hausdorffmetricχfor sets. The cardinal number of a setI is denoted by cardIandNstands for the set of the natural numbers.
With everyminNand everyC⊂Hassociate the sets Lm(C)=
x∈H: cardQ(x,C)=m, (1.2)
which are calledm-locus of the metric antiprojection generated byC.
ForX⊂HandC∈Ꮿ, the following sets are the loci generated byCinX:
LmX(C)=X∩Lm(C). (1.3)
We will also make use of the metric projection (or the nearest point mapping) defined for nonemptyM⊂Hby
P(x,M)=
y∈M:|x−y| =d(x,M), (1.4) with distance functiond(x,M)=inf{|x−z|:z∈M}.
Investigations on typical properties of antiprojections’ loci reflect the research done for metric projections (see, e.g., [3,18,20,24,26]). In [2,12], results of the following type are proved: for any bounded and closed subsetMof a uniformly convex or locally uniformly convex Banach space X, the setL1(M) is residual in X, and Q(·,M) is continuous at x∈L1(M). Zamfirescu [26] proved for metric projections that typically the complement ofL1(M) is dense. Further development for either metric projections or antiprojections is to be found in, for example, [6,7,8,9,14,28,29,30,31].
A setC⊂His calledeverywhere continualin the setX⊂Hif card(C∩X∩U)≥cfor every open setU⊂Hsuch thatX∩U = ∅. The lettercdenotes the cardinal number of the continuum.
The following main result is to be established.
Theorem1.1. LetXbe a nonempty separable convex subset of the Hilbert spaceH(Ω)and letm∈NsatisfydimX≥m−1. Then there exists a residual subsetofᏯsuch that for everyC∈the locusLmX(C)is dense in X. Moreover, ifdimX≥m, then there exists a residual setc⊂Ꮿsuch that forC∈cthe setLmX(C)contains an everywhere continual in Xsubset at each point of whichQ(·,C)is upper semicontinuous.
2. Notation
A topological spaceX is called Baire space if the intersection of every countable family of open and dense subsets ofX is dense inX. A setR⊂Xis residual if it contains some denseGδsubset ofX. IfRis residual inX, we say with some abuse of the formal logic that any elementx∈Ris a typical element ofX.
A point-to-set mappingF:X→Y, whereX and Y are topological spaces, is upper semicontinuous (u.s.c.) (resp., lower semicontinuous (l.s.c.)) atx0∈Xprovided for every open setU⊃F(x0) (resp.,U∩F(x0) = ∅), there exists an open setV ⊂X,x0∈V, such thatF(x)⊂U(resp.,F(x)∩U = ∅) wheneverx∈V.
For a subsetMofH, intM,M, spanM, coM, and diamMstand for the interior, clo- sure, span, convex hull, and diameter ofM, respectively, The set uscQM consists of all pointsx∈Hat whichQ(x,M) is upper semicontinuous. In some cases, for an arbitrary set-valued mappingF, we write uscFand lscFto denote the sets of upper semicontinuity and lower semicontinuity, respectively, ofF.
Let θ be the origin of H. The closed line segment with end-pointsx,y∈H is de- noted by [x,y]. IfMi⊂Handλi∈Rfori=1, 2, thenλ1M1+λ2M2= {λ1x1+λ2x2:x1∈ M1,x2∈M2}.
B[x,r] is the closed ball centered atx∈Hwith radiusr >0, while B(x,r) (resp., S(x,r)) is the open ball (resp., the sphere) with the same center and radius. B is the closed unit ball ofHand S is the unit sphere.
The balls in other metric spaces are denoted in a different way, that is, byB with a subscript indicating the space or without a subscript whenever there is no ambiguity. For instance,B(C,r) (resp.,B[C,r]) stands for the open (resp., closed) ball, centered atC∈Ꮿ with radiusr >0.
SupposeM1,. . .,Mmare nonempty subsets. Denote the set of equidistant points from allMiwith respect to the farthest distance by
τMi
m
i=1=
x∈H:fx,Mi
= fx,Mj
,i,j=1,. . .,m. (2.1)
Obviouslyτ(Mi)mi=1is a closed set inHand whenever nonempty, it is a complete metric space under the metric induced by| · |.
In the sequel by Hausdorfftopology (ofᏯ) we mean the topology ofᏯgenerated by the Hausdorffdistanceχ. It is well known that (Ꮿ,χ) is a complete metric space [16].
3. Topological facts
This small section contains some auxiliary results which are possibly not formulated in full generality but in a most suitable way for our purposes. The following are well-known theorems of Fort, Kuratowski and Ulam, Alexandroffand Urysohn, and Brouwer and Miranda.
Theorem3.1 (Fort [13]). SupposeXandYare complete metric spaces,Yis separable, and F:X→Y is an upper semicontinuous set-valued mapping with nonempty compact images.
ThenFis lower semicontinuous on a denseGδsubset ofX.
Theorem3.2 (Kuratowski and Ulam, see [22]). SupposeX andY are complete metric spaces,Y is separable, andRis a denseGδ subset of the product spaceX×Y. Then there exists a denseGδsubsetRX ofX such that for everyx∈RX, the set{y∈Y: (x,y)∈R}is denseGδsubset ofY.
Theorem3.3 (Alexandroffand Urysohn, see [1]). LetAbe a denseGδsubset of a nonempty metrizable compactumK, and assumeKhas no isolated points. ThencardA≥c.
Theorem3.4 (Brouwer and Miranda, see [4,21]). LetIr⊂Rnbe a bounded polyhedron of the form{x∈Rn:|vi,x| ≤r,i=1,. . .,n}wherer >0, andv1,. . .,vnare linearly indepen- dent vectors. Fori=1,. . .,n, letA±i = {x∈Ir:vi,x = ±r}and letgi:Ir→Rbe continuous
functions such thatgi(x)<0ifx∈A−i,gi(x)>0ifx∈A+i. Then there exists a pointxˆ∈Ir such thatgi( ˆx)=0for alli=1,. . .,n.
There are also several topological lemmas.
Lemma3.5 ([19], cf. [9]). SupposeX is a complete metric space andR⊂X satisfies the following property: for everyx∈X and everyη >0there are y∈X andκ >0 such that B[y,κ]⊂B(x,η)andR∩B[y,κ]is residual inB[y,κ]with respect to the relative topology.
ThenRis a residual subset ofX.
Lemma3.6 [28]. LetX andY be complete metric spaces and letF:X→Y be an upper semicontinuous set-valued mapping with nonempty compact images. Then the set
Λ=
(x,y)∈X×Y:y∈F(x),Fis l.s.c. atx (3.1) is a Baire space with respect to the relative topology induced by the topology ofX×Y.
A topological phenomenon allows us to project orthogonally residual subsets of the graph of an l.s.c. mapping onto residual subsets of its domain. The following lemma can be derived from results in [11]. A detailed proof is contained in [28].
Lemma3.7. Under the assumptions of the previous lemma, let alsoΣbe a residual subset of Λin the relative topology ofΛ. Then the orthogonal projectionπalongY mapsΣonto a residual subset ofX.
4. Lemmas
Lemma4.1. LetC0∈Ꮿ,s,r∈R,0< s < r, anda∈R. LetC0⊂B[a,s], and letY= {y1,. . ., ym}be a nonempty subset ofS(a,r). PutC=co(Y∪C0). Then for everyε∈(0,r−s)there existsδ >0such thatd(x,Y)< εwheneverx∈Cand|x−a|> r−δ.
Proof. With no loss of generality assumea=θ andr=1. Put δ=ε2/8 and take x∈ C\(1−δ)B. Denoteu=x/|x|and consider the slice Slδ(u)= {z∈B :u,z>1−δ}. In order to complete the proof it suffices to show thatd(x,Y)≥εimpliesC⊂B\Slδ(u).
Forz∈Slδ(u),
|u−z|2=1−2u,z+|z|2<2δ=ε2
4, (4.1)
that is,|u−z|< ε/2. Hence
d(z,Y)≥d(x,Y)− |x−u| − |u−z|> ε
2−δ >0, (4.2) which entails Slδ(u)∩Y = ∅. Since Slδ(u)∩B= ∅, by the choice of δ, then Slδ(u)∩ (C0∪Y)= ∅. However, the set B\Slδ(u) is closed and convex, so it contains C. The
proof is completed.
Lemma 4.2. Under the assumptions of the previous lemma for everyε∈(0,r−s) there existsδ0∈(0,ε/2)such that for everyδ∈(0,δ0], everyC∈B(C,δ), and everyx∈B(θ,δ),
the following hold:
C\Bx,f(x,C)−4δ⊂ m j=1
Byj,ε, (4.3)
∀j Byj,ε∩C\Bx,f(x,C)−4δ = ∅, (4.4) which actually means that for j=1,. . .,mthere are nonempty closed setsMj(x,C)⊂B[yj, ε]such that
C\Bx,f(x,C)−4δ= m j=1
Mj(x,C). (4.5)
Proof. There is no loss of generality in assuminga=θ. Takeε∈(0,r−s). As a conse- quence ofLemma 4.1, there existsδ>0 such that
C\
r−δB⊂ m j=1
B yj,ε 2
. (4.6)
Let nowδ0=min{δ/8,ε/2}. Takeδ∈(0,δ0],C∈B(C,δ),x∈B(θ,δ), and suppose z∈C\B(x,f(x,C)−4δ). There isu∈Csatisfying|u−z| ≤δ. It is easy to check
r−2δ < f(x,C)< r+ 2δ. (4.7) Then
r−6δ < f(x,C)−4δ≤ |z−x| ≤ |z−u|+|u|+|x|<|u|+ 2δ, (4.8) that is,|u|> r−8δ0≥r−δand (4.6) impliesd(u,Y)< ε/2.
Further,
d(z,Y)≤ |z−u|+d(u,Y)< δ+ε
2≤ε, (4.9)
which verifies (4.3).
Concerning (4.4), for eachj=1,. . .,mthere existszj∈Csuch that|zj−yj|< δ < ε.
Then by (4.7),
f(x,C)−2δ < r=yj≤yj−zj+zj−x+|x|<zj−x+ 2δ, (4.10) whence f(x,C)−4δ <|zj−x|for j=1,. . .,m. The proof is completed.
Lemma 4.3. Under the assumptions of Lemma 4.1, there existsδ0>0 such that forδ∈ (0,δ0],C∈B(C,δ),x∈B(a,δ0), andj∈ {1,. . .,m}the following inequality holds:
fx,C∩Byj,δ−x−yj≤χC,C. (4.11)
Proof. Assume without any loss of generalitya=θandr=1. Letε>0 be chosen so that the setssB and B[yj,ε] forj=1,. . .,mbe pairwise disjoint. Findγ >0 such that
Slγ
yj
=
z∈B :yj,z≥1−γ⊂Byj,ε. (4.12) Find next,ε >0 so that in turn B(yj, 2ε)⊂Slγ(yj), j=1,. . .,m.
According toLemma 4.2there existsδ0∈(0,ε/3) such that (4.3) and (4.4) hold when- everδ∈(0,δ0),C∈B(C,δ), andx∈B(θ,δ). Now, the inequality
x−yj≤fx,C∩Byj,δ+χ(C,C), j=1,. . .,m (4.13) is obviously fulfilled. In order to verify
fx,C∩Byj,δ≤x−yj+χ(C,C), j=1,. . .,m, (4.14) fix j and takey∈C∩B[yj,δ]. For everyσ∈(0,ε/3) there existsyσ∈co(C0∪Y) with
|y−yσ|< χ(C,C) +σ. Obviously,yσ∈B(yj,ε).
Our next goal is to prove
x−yσ≤x−yj. (4.15)
Ifyσ =yjandyσ is given in the form yσ=λ0y0+
m i=1
λiyi, y0∈C0, m i=0
λi=1, λi≥0,i=0,. . .,m, (4.16) then foryσ=(1−λj)−1(λ0y0+mi=1,i =jλiyi) we haveyσ∈[yj,yσ]. Ifyσ ∈B(x,f(x,C)− 4δ) then, according toLemma 4.2for the caseC=C,yσ∈ ∪mi=1B[yi,ε]. However,yσ ∈ B[yj,ε] because this ball is strongly separated from co(C0∪Y\ {yj}) by a hyperplane orthogonal toyj. Hence, there is uniquei,i =j, such thatyσ∈B[yi,ε]. In that case there existsyσ∈G= {z∈B :yi,z =γ}satisfyingyσ∈[yj,yσ]. The hyperplane segmentG∩ Cwhich is contained in B(yi,ε) is also a subset of B(x,f(x,C)−4δ), due toLemma 4.2, since it does not meet the setY+εB.
Thus, eitheryσ∈B(x,f(x,C)−4δ) or there is another pointyσfrom that ball such that yσ ∈[yj,yσ]. In both cases yσ belongs to a line segment with one end-point in B(x,f(x,C)−4δ) and the other one beingyj. On the other hand, it is easy to check
Bx,fx,C−4δ⊂Bx,x−yj, (4.17) whence (4.15) follows.
Therefore,
|x−y| ≤x−yσ+yσ−y<x−yj+χC,C+σ (4.18) for arbitraryσ∈(0,ε/2), which implies (4.14). The proof is completed.
Lemma4.4. Suppose(Cn)is a sequence inᏯ,limCn=C0, and for somey∈Handr >0, Cn∩B(y,r) = ∅forn=0, 1,. . .. Then,
limχCn∩B(y,r),C0∩B(y,r)=0. (4.19) Proof. Letε >0 be arbitrary, andσ >0 is chosen so that both σ and (σ2+ 2rσ)1/2 are smaller thanε/2. There isn0∈Nsuch that for everyn≥n0and everyx∈C0∩B(y,r) there existsxn∈Cnwith|x−xn|< σ. Ifxnis not already in B(y,r), then there arezn∈ Cn∩B(y,r) andvn∈(xn,zn)∩S(y,r).
It is easy to check|x−vn|< σ+ (σ2+ 2rσ)1/2< ε, and to find another point wn∈ (xn,zn)∩B(y,r) which is sufficiently close tovnsuch that|x−wn|< ε. Certainly,wn∈Cn. ThusCn∩B(y,r) + B(θ,ε)⊃C0∩B(y,r), whenevern≥n0. Similarly, one provesC0∩ B(y,r) + B(θ,ε)⊃Cn∩B(y,r) forn≥n0, whence
χCn∩B(y,r),C0∩B(y,r)< ε, n≥n0. (4.20)
5. Main construction
Proposition5.1. SupposeXis a nonempty separable convex subset ofH, and letm∈Nbe such that
dimX≥m. (5.1)
GivenC0∈Ꮿ,a∈X,ε >0, there areC∈Ꮿ,κ >0,
BC,κ⊂BC0,ε, (5.2)
and a residual subset(a,ε)⊂B[C,κ] with respect to the relative topology in B[C,κ]
induced by the Hausdorffmetricχsuch that
cardLmX(C)∩B(a,ε)∩uscQC≥c (5.3) wheneverC∈(a,ε), that is, them-locus ofC∈(a,ε)meets the setX∩B(a,ε)at con- tinuum points at whichQ(·,C)is upper semicontinuous.
Proof. It is no loss of generality to assume thata=θ, f(θ,C0)>2ε, and that there exists anm-dimensional subspaceHsuch thatθbelongs to the relative interior of the setX= H∩X, that is, for someε>0 the following inclusion holds:
H∩Bθ,ε⊂X. (5.4)
In casem=1, the proof is more simple and follows the scheme of the general case.
Assumem >1.
Put f0= f(θ,C0) and find pointsy1,y2,. . .,ym∈Hsuch that yj∈S θ,f0+ε
3
, dyj,C0
<ε
2, j=1,. . .,m. (5.5)
Certainly,
dyj,C0
≥ ε
3, j=1,. . .,m. (5.6)
Denote byπthe orthogonal projection ontoH. LetY= {y1,. . .,ym}andvj=π(yj+1−y1) forj=1,. . .,m−1. Takev0∈H\spanV and designate
V=
v1,. . .,vm−1
, V= v0
∪V. (5.7)
It is assumed also, which is no loss of generality, that the elements ofYhave already been chosen in such a way that the elements ofV are linearly independent, that is, spanV =H.
Put
C=coC0∪Y. (5.8)
According toLemma 4.3there existsε>0 such that for everyδ∈(0,ε],C∈B[C,δ], x∈B(θ,δ), and every j=1,. . .,mthe inequality (4.11) holds.
Denoteε=4−1min{|yi−yj|:i,j=1,. . .,m,i =j}and ε0=min
ε,ε,ε,ε 6
. (5.9)
In view ofLemma 4.2there exists a numberδ, from now on it is fixed, 0< δ <ε0
4, (5.10)
such that for everyC∈B(C,δ) andx∈B(θ,δ) there are nonempty closed setsMj(x,C) forj=1,. . .,msuch that
C\Bx,f(x,C)−4δ= k j=1
Mj(x,C), Mj(x,C)⊂Byj,ε0
. (5.11)
Now forr >0 defineIr= {x∈H:|v,x| ≤r,v∈V}.SinceV is a basis forH, then for eachr >0 the setIr is bounded and lim diamIr=0. Having also in mind (5.4), fix r∈Rsuch that
Ir⊂X∩B(θ,δ). (5.12)
The following functions are defined onH:
γj(x)=x−y1−x−yj+1, j=1,. . .,m−1. (5.13) Denoting
A±j =
x∈Ir:vj,x= ±r, j=1,. . .,m−1, (5.14)
observe that for any j=1,. . .,m−1 the sign ofγj(·) is equal to the sign of the inner productvj,·. This is a consequence of the equalities
x−y12−x−yj+12=2yj+1−y1,x=
vj,x, (5.15) since
yj+1−y1−vj,x=0 forx∈H, j=1,. . .,m−1. (5.16) Thusγj(·) take opposite sign values on the corresponding facesA−j andA+j.
SinceA±j are compact,γj(x) attains minimal and maximal values on them for each j=1,. . .,m−1. Denote forj=1,. . .,m−1,
−αj=maxγj(x) :x∈A−j, βj=minγj(x) :x∈A+j, (5.17) Chooseκ >0 such that
κ < δ, (5.18)
κ≤1
3minαj,βj:j=1,. . .,m−1. (5.19) By (5.8), (5.5), (5.10), and (5.18), forC∈B[C,κ] we have
χC,C0
≤χC,C+χC,C0
< κ+ ε
2< ε, (5.20)
thus verifying the inclusion (5.2).
For every C∈B[C,κ] and j=1,. . .,m, define Nj(C)=C∩B[yj,ε0]. Notice that (4.11) can be rewritten in the following way:
fx,Nj(C)−x−yj≤κ, j=1,. . .,m. (5.21) Consider the following functions defined onH∩B(θ,δ):
γj(x,C)=fx,N1(C)−fx,Nj+1(C), j=1,. . .,m−1. (5.22) Lett∈[−r,r] be fixed. DenotingLt= {x∈H:v0,x =t}, andIr(t)=Ir∩Lt, we ap- ply the Brouwer-Miranda theorem to the setIr(t) and the functions defined by (5.22). In order to verify the boundary, conditions take, for instance,x∈A−j ∩Ltforj=1,. . .,m− 1, and then, having in mind (5.19) and (5.21),
γi(x,C)≤x−y1−x−yi+1+ 2κ≤ −αi+ 2κ <0. (5.23) Analogously, forx∈A+i ∩Lt,
γi(x,C)≥x−y1−x−yi+1−2κ≥βi−2κ >0. (5.24)
Therefore, for everyt∈[−r,r] there exists at least one point ˆxt∈Ir(t) at which all functions in (5.22) vanish, that is, ˆxt∈τ(Nj(C))mj=1. Now, according to (5.11) and (5.18) for everyx∈B(θ,δ) andC∈B[C,κ], we haveMj(x,C)⊂Nj(C), j=1,. . .,m, and then
Q(x,C)⊂ m j=1
Nj(C), (5.25)
Thus for everyC∈B[C,κ] andt∈[−r,r] there is ˆxt∈Ir(t) such that forj=1,. . .,m, fxˆt,C=fxˆt,Nj(C). (5.26) In order to prove that the antiprojectionsQ(·,C) are actuallym-valued and u.s.c. at
“many points aroundθ,” we make further considerations involving topological lemmas fromSection 3. Introduce complete metric spaces᐀=B[C,κ]×[−r,r] and=᐀×Ir
with the box metricρon the products, that is, for (C,t,x), (D,s,y)∈, ρ(C,t,x), (D,s,y)=maxχ(C,D),|t−s|,|x−y|
. (5.27)
Define a set-valued mappingG:᐀→Irby
G(C,t)=Ir(t)∩τNj(C)mj=1. (5.28) The images ofGare nonempty, and by the continuity of the farthest distance function they are closed, hence compact asIris compact.
Also, in view ofLemma 4.4,Gis upper semicontinuous. To verify this, it is sufficient to notice that if (Cn,tn) is a sequence in᐀convergent to some (C,t)∈᐀, then the se- quences (Nj(Cn)) converge toNj(C) forj=1,. . .,mwith respect to the distanceχ, while Ir(tn) obviously converges toIr(t). Having in mind the continuity of the farthest distance f(·,·) with respect to both the arguments and the compactness ofIr, it is possible for any sequence (un),un∈G(Cn,tn) to find a cluster point inG(C,t).
The theorem of Fort implies that lscGis a residual subset of᐀. Denote byΛthe graph ofG
Λ=
(C,t,x)∈:x∈G(C,t), (5.29)
and byΛthe “graph of lower semicontinuity”
Λ=
(C,t,x)∈Λ: (C,t)∈lscG. (5.30) According toLemma 3.6,Λis a Baire space in the relative topology induced by the prod- uct topology of᐀×Ir.
Let forn∈N, ᐁn=
(C,t,x)∈Λ:∃s, 0< s <4δ,
diamNj(C)\Bx,fx,Nj(C)−s< n−1, j=1,. . .,m. (5.31)
Claim. ᐁkcontains open dense set inΛ.
Take arbitrary (D0,t0,x0)∈Λandσ >0. It has to be proved thatB((D0,t0,x0),σ)∩ ᐁn contains an open subset of Λ. Assume without any loss of generality that D0∈ B(C,κ). Indeed, due to the l.s.c. ofGat (D0,t0) and, by [13], the fact that lscGis residual in the open dense subsetB(C,κ)×[−r,r] of᐀, one can conclude that arbitrary close to (D0,t0,x0) there are points (D,t,x) fromΛwithD∈B(C,κ). Hence there isλ∈(0,κ), λ < σ/2, such thatχ(D0,C)< κ−λ.
DenotingKj=(H\B[x0,f(x0,D0)])∩B(yj,ε0), j=1,. . .,m, we haveKj = ∅. Cer- tainly, there are y∈B(yj,ε0) satisfying|y|>|yj|+ 3ε0/4 and if we assume |x0−y| ≤
f(x0,D0) for ally∈B(yj,ε0), then (also having in mind (5.10) and (5.18))
|y| ≤y−x0+x0< fx0,D0
+δ <yj+κ+ 2δ <yj+3ε0
4 (5.32)
gives a contradiction.
Since
infdz,Nj D0
:z∈Kj
=0, j=1,. . .,m, (5.33) it is a matter of routine to find pointszj∈Kj,j=1,. . .,m, andξ >0 such that
dzj,Nj
D0
< λ, zj∈Sx0,fx0,D0
+ξ, j=1,. . .,m. (5.34) Consider the setD=co(D0∪ {z1,. . .,zm}). Obviously,Q(x0,D)= {z1,. . .,zm}.
Now, letµ >0,µ <min{ξ,λ, (2n)−1}be chosen so that B(zj,µ)⊂B(yj,ε0) for all j= 1,. . .,m. ApplyLemma 4.2with respect to the setsD0,Dand the pointx0(instead ofC0, Candθ, resp.). There isη∈(0,µ/2) such that for everyD∈B[D,η] andx∈B(x0,η) there are nonempty closed setsMj(x,D)⊂B[zj,µ], j=1,. . .,m, satisfying
D\Bx,f(x,D)−4η= m j=1
Mj(x,D). (5.35)
Notice that diamMj(x,D)≤2µ < n−1and then fors=4η,
diamNj(D)\Bx,fx,Nj(D)−s< n−1, j=1,. . .,m. (5.36) AlsoB[D,η]⊂B[C,κ], and the claim easily follows from the lower semicontinuity ofG applied forx0∈G(D0,t0).
Finally, putᐁ= ∩∞n=1ᐁn. The setᐁis residual inΛand byLemma 3.7is orthogonally projected on a residual subsetᐂof ᐀. Thus for every (C,t)∈ᐂthere isx(t)∈G(C,t) such that
x(t)∈Ir(t)∩τNj(C)mj=1, (5.37) and all mappingsQ(·,Nj(C)) forj=1,. . .,mare single-valued and upper semicontinu- ous atx(t). Apply the Kuratowski-Ulam theorem to the product space᐀to show the ex- istence of a residual subset(θ,ε) ofB[C,κ] such that, in view of Alexandroff-Urysohn
theorem, (5.37) is satisfied for a continuum of realst∈[−r,r], wheneverC∈(θ,ε).
Thus (5.3) is verified and the proof is completed.
Proposition5.2. If inProposition 5.1the condition (5.1) is replaced by the following one:
dimX=m−1, (5.38)
then in the conclusion (5.3) is to be replaced by
CXm(C)∩B(a,ε)∩uscQC = ∅ forC∈(a,ε). (5.39) Proof. In casem=1, the setXis a singleton and it is a matter of routine to have a direct proof of the fact that there exists a residual subset of a ballB[C,κ]⊂Ꮿsuch that both (5.2) and (5.39) are fulfilled.
The proof in the general case differs slightly from the proof of the previous propo- sition. For instance, the theorem of Brouwer-Miranda is applied with respect to the set Ir= {x∈H:|v,x| ≤r,v∈V}instead ofIr, andH=spanV. 6. Proof ofTheorem 1.1
Suppose dimH≥m. ByProposition 5.1 and byLemma 3.5for everya∈X and every ε >0, there is a residual subset(a,ε) ofᏮsuch that
cardCn,mX (C)∩B(a,ε)∩uscQC
≥c forC∈(a,ε). (6.1) Let{a1,a2,. . .}be a countable dense set inX. Put
= ∞ n=1
an,n−1. (6.2)
For everyC∈the locusLmX(C) intersects an arbitrary nonempty open subsetU of Xat a set containing at least continuum points of upper semicontinuity ofQ(·,C), that is, the setLmX(C)∩uscQCis everywhere continual inX.
If dimH=m−1, then byProposition 5.2andLemma 3.5again, for typicalC∈Ꮿthe setLmX(C)∩uscQCis dense inX. The theorem is proved.
Corollary6.1. IfX is a nonempty convex subset ofH(Ω)andΩis a finite set, that is, H(Ω)=Rk,k∈N, then for a typical convex compactKofRkthe lociLmX(K)inXpartition Xinto a finite sequence of dense sets such that
(i)L1X(K)is denseGδ,
(ii)LmX(K)are everywhere continual, for1< m≤dimX, (iii)LmX(K)are dense, wheneverm=dimX+ 1.
Proof. (i) Ifm=1 and dimX≥1, then the mappingQ(·,M) is u.s.c. with nonempty images, and it is single-valued on a dense subset ofX. It follows from a known result, for instance [29], that it is single-valued on a denseGδsubset ofX.
To prove (ii) and (iii), recall a result from [14], proved in another setting but adaptable for the present purpose, which states thatLmX(K)= ∅for typicalK∈Ꮿwheneverm >
dimX+ 1. ApplyProposition 5.1for establishing (ii), andProposition 5.2for (iii).
Corollary6.2. SupposeΩis a countable set, that is,H(Ω)is a separable Hilbert space.
Then for everym∈Nthere exists a residual subsetm ofᏯsuch thatLm(C)is dense in H(Ω)wheneverC∈m.
Acknowledgment
The research of the second author is in part supported by G.N.A.F.A., Italy.
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